How Do You Calculate Building Height Using Free Fall Formulas with Given Times?

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SUMMARY

The discussion focuses on calculating the height of a building using free fall formulas, specifically with given times of 3.25 seconds for acceleration and 1.25 seconds for deceleration. The primary formula utilized is Dy = V0t + 1/2at², where V0 is the initial velocity, which is zero in this case. The calculated height, using the acceleration due to gravity (9.8 m/s²), results in a height of 51.8 meters. Participants express confusion regarding the application of the formula and the overall problem setup.

PREREQUISITES
  • Understanding of kinematic equations, specifically Dy = V0t + 1/2at²
  • Knowledge of gravitational acceleration (9.8 m/s²)
  • Basic algebra skills for manipulating equations
  • Familiarity with concepts of free fall and motion under gravity
NEXT STEPS
  • Study the derivation and application of kinematic equations in physics
  • Learn about the effects of acceleration and deceleration on free fall
  • Explore real-world applications of free fall calculations in engineering
  • Investigate the impact of air resistance on falling objects
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding motion under gravity and free fall calculations.

PhunPhysicsss
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Homework Statement


Figure out the height of the building using free fall formulas and only knowing two times: 3.25 second acceleration, 1.25 seconds decceleration. Find the height.

Homework Equations


Dy=V0t + 1/2at2 (only one I can think of that applies)

The Attempt at a Solution



Dy=V0t + 1/2at2
V0t=0

Dy= 1/2at2

Dy= .5(9.8m/s2)(10.5625)

Dy= 51.8 m

I am really stumped on this for some reason! I'm not even sure if I am using the correct formula. Any and all help is appreciated! Thanks y'all!
 
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Welcome to PF!

Hi PhunPhysicsss! Welcome to PF! :wink:
PhunPhysicsss said:
Figure out the height of the building using free fall formulas and only knowing two times: 3.25 second acceleration, 1.25 seconds decceleration. Find the height.

Sorry, I've no idea what you mean. :redface:

What's the full question? :smile:
 

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